construction of the scattering states, $\psi_p(x)$ incident from the left, $\psi'_p(x)$ incident from the right
\begin{eqnarray}
\psi_p(x)=\left\{\begin{array}{cc}
e^{ipx}+r(p)e^{-ipx},&x\rightarrow-\infty\\
te^{ipx},&x\rightarrow+\infty
\end{array}\right.
\end{eqnarray}
\begin{eqnarray}
\psi'_p(x)=\left\{\begin{array}{cc}
e^{-ipx}+r'(p)e^{-ipx},&x\rightarrow+\infty\\
t'(p)e^{ipx},&x\rightarrow -\infty
\end{array}\right.
\end{eqnarray}
<sub>I write $x\rightarrow\pm\infty$, since it is not enough to take $x$ outside of the support of $V$, because of evanescent waves: waves that decay exponentially into the region where $V=0$, but have not yet decayed to zero.</sub>  

the reflection coefficients $r(p),r'(p)$ and transmission coefficients $t(p),t'(p)$ define the scattering matrix:

$$S(p)=\begin{pmatrix}
r(p)&t'(p) \\
t(p)&r'(p)
\end{pmatrix}$$
unitarity: $S(p)S^\dagger(p)=\mathbb{1}$, time-reversal symmetry: $S^t(p)=S(p)$, so $t(p)=t'(p)$.